Question

What is the hybridization of copper in \([Cu(NH_3)_4]^{2+}\)?

• A. \(sp^3\)
• B. \(sp^2d\)
• C. \(dsp^2\)
• D. \(sp^3d\)

Hint:

First work out the oxidation state and d-electron count of copper, then see how many and which orbitals are needed for 4 ligands.

Answer 1

The Correct Option is C

Step 1: Find the oxidation state of copper. The complex ion carries an overall charge of 2+, and ammonia is a neutral ligand, so all four \(NH_3\) molecules together contribute zero charge. This means copper itself must carry the +2 charge, so we are dealing with \(Cu^{2+}\).
Step 2: Write the electron configuration of \(Cu^{2+}\). Neutral copper is \([Ar]3d^{10}4s^1\). To form \(Cu^{2+}\), we remove 2 electrons, first the single 4s electron and then one 3d electron, giving \[Cu^{2+} : [Ar]3d^9\]
Step 3: With a \(d^9\) configuration, one of the five 3d orbitals is singly occupied. In forming this four coordinate ammine complex, copper uses one empty (or vacated) 3d orbital along with the 4s orbital and two 4p orbitals to accept the four lone pairs from the nitrogen atoms of ammonia. This combination of one d, one s and two p orbitals is \(dsp^2\) hybridization, which gives a square planar arrangement of the four ligands around copper.
\[\boxed{dsp^2 \text{ hybridization, square planar shape}}\]